Leetcode: Isomorphic Strings

Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

For example,
Given "egg", "add", return true.

Given "foo", "bar", return false.

Given "paper", "title", return true.

Note:
You may assume both s and t have the same length.

Best: O(N) time, constant space

 1 public class Solution {
 2     public boolean isIsomorphic(String s1, String s2) {
 3         int[] m = new int[256];
 4         int[] n = new int[256];
 5         for (int i = 0; i < s1.length(); i++) {
 6             if (m[s1.charAt(i)] != n[s2.charAt(i)]) return false;
 7             m[s1.charAt(i)] = n[s2.charAt(i)] = i+1;
 8         }
 9         return true;
10     }
11 }

复杂度

时间 O(N) 空间 O(N)

思路

用一个哈希表记录字符串s中字母到字符串t中字母的映射关系，一个集合记录已经映射过的字母。或者用两个哈希表记录双向的映射关系。这里不能只用一个哈希表，因为要排除egg->ddd这种多对一的映射。

 1 public class Solution {
 2     public boolean isIsomorphic(String s, String t) {
 3         Map<Character, Character> map = new HashMap<Character, Character>();
 4         Set<Character> set = new HashSet<Character>();
 5         if(s.length() != t.length()) return false;
 6         for(int i = 0; i < s.length(); i++){
 7             char sc = s.charAt(i), tc = t.charAt(i);
 8             if(map.containsKey(sc)){
 9                 // 如果已经给s中的字符建立了映射，检查该映射是否和当前t中字符相同
10                 if(tc != map.get(sc)) return false;
11             } else {
12                 // 如果已经给t中的字符建立了映射，就返回假，因为出现了多对一映射
13                 if(set.contains(tc)) return false;
14                 map.put(sc, tc);
15                 set.add(tc);
16             }
17         }
18         return true;
19     }
20 }